Question

Two moles of $P{Cl}_5$ were introduced in a $2$ litre flask and heated at $600K$ to attain equilibrium. $P{Cl}_5$ was found to be $40$% dissociated into $P{Cl}_3$, ${Cl}_2$. Calculate the value of $K_c$.

Answer 1

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
Initially 2 moles of $PC{l}_5$, 0 moles of $PC{l}_3$ and 0 moles of $C{l}_2$ were present.
To reach equilibrium, 40 % of 2 moles or $\frac{\text{40}}{\text{100}}\times 2=0.8$ moles of $PC{l}_5$
dissociate to form 0.8 moles of $PC{l}_3$ and 0.8 moles of
$C{l}_2$.
$2-0.8=1.2$ moles of $PC{l}_5$ remains.
The equiibrium concentrations are
$\left[PC{l}_5\right]=\frac{\text{1.2 mol}}{\text{2 L}}=\text{0.6 M}$
$\left[PC{l}_3\right]=\frac{\text{0.8 mol}}{\text{2 L}}=\text{0.4 M}$
$\left[C{l}_2\right]=\frac{\text{0.8 mol}}{\text{2 L}}=\text{0.4 M}$
The equilibrium constant
$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$
$K_c=\frac{\text{0.4 M}\times \text{0.4 M}}{\text{0.6 M}}$
$K_c=0.267mol/L$